The absolute maximum value of y = x3 − 3x + 2, 0 ≤ x ≤ 2, is

The correct answer is Option (1) → 4

On a closed interval, the absolute maximum is the largest of the values at the critical points and the end points.

    \(y' = 3x^2 - 3 = 0\)
    \(x = \pm 1\)

Only x = 1 lies in [0, 2]. The values:

    \(y(0) = 2\)
    \(y(1) = 1 - 3 + 2 = 0\)
    \(y(2) = 8 - 6 + 2 = 4\)

The absolute maximum is 4, at x = 2.